If $g(x)=\left(x^2+2 x+3\right) f(x), f(0)=5$ and $\lim\limits_{x \rightarrow 0} \frac{f(x)-5}{x}=4$, then g'(0) is equal to

22

We have,

$f(0)=5$  and  $\lim\limits_{x \rightarrow 0} \frac{f(x)-5}{x}=4$

∴  $\lim\limits_{x \rightarrow 0} \frac{f(x)-5}{x}=4$

$\Rightarrow \lim\limits_{x \rightarrow 0} \frac{f(x)-f(0)}{x}=4 \Rightarrow f'(0)=4$

Now,

$g(x)=\left(x^2+2 x+3\right) f(x)$

$\Rightarrow g'(x)=(2 x+2) f(x)+\left(x^2+2 x+3\right) f'(x)$

$\Rightarrow g'(0)=2 f(0)+3 f'(0)=2 \times 5+3 \times 4=22$